An open lift is moving upward with velocity `10 m//s.` It has an upward acceleration of `2 m//s^2.` A ball is projected upwards with velocity `20 m//s` relative to ground. Find
(a) time when ball again meets the lift
(b) displacement of lift and ball at that instant.
(c) distance travelled by the ball upto that instant.
Take `g=10 m//s^2`

Explanation

(i) At the time when ball again meets the lift, `s_(L)=s_(B)`
`therefore 10 t+(1)/(2)xx2xxt^(2)=20t-(1)/(2)xx10t^(2)`
Solving this equation, we get

`therefore` Ball will again meet the lift after `(5)/(3)s`.
(ii) At this instant `s_(L)=s_(B)=10xx(5)/(3)+(1)/(2)xx2xx((5)/(3))^(2)`
`=(175)/(9)m=19.4 m`
(iii) For the ball `u uarr darr a`. Therefore we will first find `t_(0)`, the time when its velocity becomes zero.
`t_(0)=|(u)/(a)|=(20)/(10)=2s`
As `t(=(5)/(3)s)lt t_(0)`, distance and displacement are equal
or `d = 19.4 m`
Concept of relative motions is more useful in two body problem in two (or three) dimensional motion. This can be understood by the following example.